# Kinematics question projectiles with vector components (baseball player Q)

• chiuda
In summary, the baseball player makes perfect contact with a ball striking it at a 45° angle above the horizontal at a point 1.3 m above the ground. His ball just makes it over the wall, traveling at 217 m/s (horizontal) and 307 m/s (vertical).
chiuda

## Homework Statement

Base ball player makes perfect contact with a ball striking it at 45° angle above the horizontal at a point 1.3 m above the ground. His ball just makes it clear of the 3m wall 130m from home plate. What was the velocity at with he struck the ball.

## Homework Equations

is my final answer correct? i think my steps are good and that my assumptions are perfectly reasonable but i am getting a very unrealistic answer, please can somebody help me it would be very appreciated?! :)

## The Attempt at a Solution

Ok so i assumed that when the question states the ball just makes it over the wall that its vertical velocity is 0m/s then assuming that i plugged it into this kinematic equation
d=(v2)(t)- (1/2)(a)(t)^2 (note i am trying to find time just through the vertical components) i then subtracted the height of which the ball was hit with where v2 was 0m/s which is 3m giving me 1.7 m for vert displacement so i have d i have v2 and i have a with is -9.8m/s^2. after plugging this into the equation and solving for t i get 0.6s. i then move on to find the horizontal velocity since it is constant and i now have time so i plug it into v=d/t because i have d horizontal (130m) and t (0.6s) from this i get 217 m/s (v horizontal) i then automatically know vi= 217 m/s as well because it was projected at a 45° angle then i use pathagreans theorem and solve for Vi and end up with 307 m/s [45°].

chiuda said:
Ok so i assumed that when the question states the ball just makes it over the wall that its vertical velocity is 0m/s then assuming that i plugged it into this kinematic equation

I would think it means that at 130 m the ball was 3 m above the ground. It's vertical speed won't be zero until it hits the ground (and bounces a few times, but that can be ignored).

yes i understand that but when the ball reaches its maximum height it has a vertical velocity which I'm assuming that the 3 m above ground is its maximum height

3 m above ground is NOT the maximum height. That is only the height of the fence, the ball needs to be hit much higher if it is to clear the fence. And it should be obvious that the ball could not be traveling at over 1000km/h initially...

Hint: What you need to find out is the t, perhaps in a form involving another variable.

I know that it's unrealistic that's why I asked the question I know I need to find t but it's unknown and I don't have enough information to find it! Ahhhhhhhhhh!

Like I said... t will end up containing another variable. That's okay, because it will cancel out in the end. t = $\frac{d}{v}$. What is your horizontal d and horizontal v?

i only have dh no vh or t my d horizontal is 130 m

Again... Stop trying to plug in numbers for now. What is your horizontal velocity in terms of variables only?

horiz velocity is d

sorry horiz velocity is v

horiz displacement is d

chiuda said:
sorry horiz velocity is v

No... v is your velocity at 45°. What is your horizontal velocity, if the velocity at 45° is v?

Hint: Use the Pythagorean theorem.

Ok horizon velocity is Vh

... NO. If my hypotenuse is v, then my horizontal component of my v must equal vcos45. That's what you need to plug in.

OK. I don't have the horizontal component of the initial velocity

$x=u_xt$
$y=u_yt-\frac{1}{2}at^2$

$u_x=uCos45°$
$u_y=uSin45°$

At time t, the ball is 130 from base(x) and at height of 3m(y)
You have 2 equations,2 unknowns.

OMG I JUST DID IT WITH A FRIEND THANK YOU SO MUCH! Sorry for the confusion, but that one was a doozy

## 1. How do you calculate the initial velocity of a projectile?

The initial velocity of a projectile can be calculated by using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity, and t is the time the projectile has been in motion.

## 2. What is the formula for calculating the horizontal distance traveled by a projectile?

The formula for calculating horizontal distance is d = ut + 1/2at^2, where d is the distance traveled, u is the initial horizontal velocity, a is the acceleration due to gravity, and t is the time the projectile has been in motion.

## 3. Can you explain the concept of vector components in kinematics?

Vector components refer to the breaking down of a vector into its horizontal and vertical components. In kinematics, this is often used to analyze the motion of a projectile, with the horizontal component representing the projectile's motion in the x-axis and the vertical component representing its motion in the y-axis.

## 4. How do you calculate the maximum height reached by a projectile?

The maximum height reached by a projectile can be calculated using the formula h = (u^2*sin^2(theta))/2g, where h is the maximum height, u is the initial velocity, theta is the angle of projection, and g is the acceleration due to gravity.

## 5. What is the difference between displacement and distance in kinematics?

In kinematics, displacement refers to the change in an object's position from its initial position to its final position. It is a vector quantity and takes into account the direction of motion. Distance, on the other hand, refers to the total length of the path traveled by an object, regardless of its direction. It is a scalar quantity and does not take into account direction.

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